For the equation 3x^2 + px + 3 = 0, find the value(s) of p if one root is (i) square of the other (ii) fourth power of the other.


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Question Number 20670 by Tinkutara last updated on 31/Aug/17

$F o r t h e e q u a t i o n 3 x^{2} + p x + 3 = 0,$ $f i n d t h e v a l u e (s) o f p i f o n e r o o t i s$ $(i) s q u a r e o f t h e o t h e r$ $(i i) f o u r t h p o w e r o f t h e o t h e r .$
	Answered by dioph last updated on 31/Aug/17
	$(i) x_1 = x_2 ^2 1 = x_1 x_2 = x_2 ^3 ∴ x_2 = 1^(1/3) , x_1 = 1^(2/3) p = −3(x_1 +x_2 )=−3×1^(1/3) −3×1^(2/3) values of p: let ω_3 = e^(i2π/3) = (−0.5+i((√3)/2)) { ((−3×1−3×1 = −6)),((−3×ω_3 −3×ω_3 ^2 = 3)) :} (ii) x_1 = x_2 ^4 1 = x_1 x_2 = x_2 ^5 ∴ x_2 = 1^(1/5) , x_1 = 1^(4/5) p = −3(x_1 +x_2 )=−3×1^(1/5) −3×1^(4/5) values of p: let ω_5 = e^(i2π/5) = (cos ((2π)/5)+isin ((2π)/5)) if cos ((2π)/5) = (((√5)−1)/4) (from regular penthagon): { ((−3×1−3×1 = −6)),((−3×ω_5 −3×ω_5 ^4 = −6 cos ((2π)/5) = −((3((√5)−1))/2))),((−3×ω_5 ^2 −3ω_5 ^3 = 6 cos (π/5) = 3 + ((3((√5)−1))/2) )) :}$
	$(i) x_{1} = x_{2}^{2}$ $1 = x_{1} x_{2} = x_{2}^{3}$ $∴ x_{2} = 1^{1 / 3}, x_{1} = 1^{2 / 3}$ $p = - 3 (x_{1} + x_{2}) = - 3 \times 1^{1 / 3} - 3 \times 1^{2 / 3}$ $values of p :$ $let ω_{3} = e^{i 2 π / 3} = (- 0.5 + i \frac{\sqrt{3}}{2})$ ${\begin{cases} - 3 \times 1 - 3 \times 1 = - 6 \\ - 3 \times ω_{3} - 3 \times ω_{3}^{2} = 3 \end{cases}$ $(i i) x_{1} = x_{2}^{4}$ $1 = x_{1} x_{2} = x_{2}^{5}$ $∴ x_{2} = 1^{1 / 5}, x_{1} = 1^{4 / 5}$ $p = - 3 (x_{1} + x_{2}) = - 3 \times 1^{1 / 5} - 3 \times 1^{4 / 5}$ $values of p :$ $let ω_{5} = e^{i 2 π / 5} = (\cos \frac{2 π}{5} + i \sin \frac{2 π}{5})$ $if \cos \frac{2 π}{5} = \frac{\sqrt{5} - 1}{4} (from regular penthagon) :$ ${\begin{cases} - 3 \times 1 - 3 \times 1 = - 6 \\ - 3 \times ω_{5} - 3 \times ω_{5}^{4} = - 6 \cos \frac{2 π}{5} = - \frac{3 (\sqrt{5} - 1)}{2} \\ - 3 \times ω_{5}^{2} - 3 ω_{5}^{3} = 6 \cos \frac{π}{5} = 3 + \frac{3 (\sqrt{5} - 1)}{2} \end{cases}$
	Commented by Tinkutara last updated on 31/Aug/17

	$Thanks for (i) part .$ $But in (i i) part, answer is 0 and$ $\frac{3 (\sqrt{5} - 1)}{2} . Can you get it ?$
	Commented by dioph last updated on 31/Aug/17

	$I added the substitution$ $\cos \frac{2 π}{5} = \frac{\sqrt{5} - 1}{4}$ $but I cannot see how to get these answers$ $If p = 0 then roots are i and - i,$ $which are not related by a$ $fourth power$
	Commented by Tinkutara last updated on 31/Aug/17

	$Thank you very much Sir!$
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